Question

Решите систему 2x^2-3xy+2y^2=4 ; 2x^2+3y^2=14

Answer 1

$\left \{ {{2x^2-3xy+2y^2=4\; |\cdot 14} \atop {2x^2+3y^2=14\; |\cdo (-4)}} \right. \oplus \left \{ {{2x^2-3xy+2y^2=4} \atop {20x^2-42xy+16y^2=0}} \right. \\\\20x^2-42xy+16y^2=0\; |:2\\\\10x^2-21xy+8y^2=0\; |:y^2\ne 0\\\\10\cdot (\frac{x}{y})^2-21\cdot \frac{x}{y}+8=0\\\\t=\frac{x}{y}\; :\; \; 10t^2-21t+8=0\; ,\; \; D=21^2-4\cdot 10\cdot 8=121\\\\t_1=\frac{21-11}{20}=\frac{1}{2}\; ,\; \; t_2=\frac{21+11}{20}=\frac{8}{5}\\\\a)\; \; \frac{x}{y}=\frac{1}{2}\; \; \to \; \; 2x=y$

$2x^2+3y^2=2x^2+3\cdot (2x)^2=14x^2\; ,\; \; 14x^2=14\; ,\; \; x^2=1\; ,\\\\x=\pm 1\; \; \to \; \; y=\pm 2\\\\b)\; \; \frac{x}{y}=\frac{8}{5}\; ,\; \; y=\frac{5x}{8}\\\\2x^2+3y^2=2x^2+3\cdot (\frac{5x}{8})^2=2x^2+\frac{75x^2}{64}=\frac{203x^2}{64}\; ,\; \; \frac{203x^2}{64}=14\; ,\\\\x^2=\frac{896}{203} =\frac{128}{29}\; ,\; \; x=\pm \sqrt{\frac{128}{29}}=\pm 8\cdot \sqrt{\frac{2}{29}}\; \; \to \; \; y=\pm \frac{5}{8}\cdot \sqrt{\frac{128}{29}}=\pm 5\cdot \sqrt{\frac{2}{29}}$

$Otvet:\; \; (-1\, ,\, -2)\; ,\; \; (1\; ,\; 2)\; ,\; \Big (-8\cdot \sqrt{\frac{2}{29}}\, ,\, -5\cdot \sqrt{\frac{2}{29}}\Big )\; ,\\\\\Big (8\cdot \sqrt{\frac{2}{29}}\, ,\, 5\cdot \sqrt{\frac{2}{29}}\Big )\; .$